What are the foci of the hyperbola with equation 16y^2-9x^2=144
why cant you just give the answer?!?!?!?!?!?
Answer is (0plus or minus 5)
16y + 0 = 16y
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To find the foci of a hyperbola, we need to focus on the equation of the hyperbola: 16y^2 - 9x^2 = 144. I'll explain step-by-step how to find the foci.
Step 1: Put the equation in standard form
The standard form of a hyperbola equation is:
[(y - k)^2 / a^2] - [(x - h)^2 / b^2] = 1 (for a horizontal hyperbola)
[(x - h)^2 / a^2] - [(y - k)^2 / b^2] = 1 (for a vertical hyperbola)
In our equation, 16y^2 - 9x^2 = 144, it's evident that the term with the negative sign is associated with x^2, so it's a horizontal hyperbola. Dividing every term by 144, we get:
(y^2 / 9) - (x^2 / 16) = 1
Step 2: Identify the values of a, b, and (h, k)
Comparing our equation with the standard form, we can see that a = 3 (sqrt of 9) and b = 4 (sqrt of 16). The values h and k can be found by moving the constant term to the right side of the equation:
(y^2 / 9) = (x^2 / 16) + 1
Now, h = 0 and k = 0 because x and y terms are not shifted.
Step 3: Calculate c
To find c, which represents the distance from the center of the hyperbola to the foci, we can use the formula c = sqrt(a^2 + b^2).
In our case, c = sqrt(9 + 16) = 5.
Step 4: Determine the foci
The foci lie on the transverse axis (the horizontal axis for a horizontal hyperbola) and are equidistant from the center, which in this case is at (h, k) = (0, 0). The foci are located (±c, 0) units away from the center.
Therefore, the coordinates of the foci are (-5, 0) and (5, 0).
y^2/9 - x^2/16 = 1
so, what are the foci of y^2/a^2 - x^2/b^2 = 1?